Question

please send me the answer​

Answer 1

Formula :

  If three terms a, b, c be in AP, we can say

                        a - b = b - c

Proof :

Given that, a², b², c² are in AP. Then,

              a² - b² = b² - c²

(i) Now, a² - b² = b² - c²

  or, (a - b) (a + b) = (b - c) (b + c)

  or, $\frac{a - b}{b + c} = \frac{b - c}{a + b}$

  or, $\frac{c + a - b - c} {(b + c) (c + a)}= \frac{a + b - c - a} {(c + a) (a + b)}$

  or, $\frac{(c + a) - (b + c)} {(b + c) (c + a)}= \frac{(a + b) - (c + a)} {(c + a) (a + b)}$

or, $\boxed{\frac{1}{b + c}- \frac{1}{c + a} = \frac{1}{c + a}- \frac{1}{a + b}}$

  which shows that

  $\frac{1}{b + c}, \frac{1}{c + a} , \frac{1}{a + b}$ are in AP.

(ii) Now, a² - b² = b² - c²

  or, (a - b) (a + b) = (b - c) (b + c)

  or, $\frac{a - b}{b + c}= \frac{b - c}{a + b}$

  or, $\frac{(a - b) (a + b + c)} {b + c}= \frac{(b - c) (a + b + c)}{a + b}$

  or, $\frac{(a - b) (a + b) + c (a - b)} {(b + c) (c + a)}= \frac{(b - c) (b + c) + a (b - c)} {(a + b) (c + a)}$

  or, $\frac{a^{2}- b^{2} + ca - bc} {(b + c) (c + a)}= \frac{b^{2} - c^{2} + ab - ca} {(a + b) (c + a)}$

  or, $\frac{(ca + a^{2}) - (b^{2} + bc)}{(b + c) (c + a)}= \frac{(ab + b^{2}) - (c^{2} + ca)}{(a + b) (c + a)}$

  or, $\frac{a (c + a) - b (b + c)}{(b + c) (c + a)}= \frac{b (a + b) - c (c + a)} {(a + b) (c + a)}$

or, $\boxed{\frac{a}{b + c} - \frac{b}{c + a}= \frac{b}{c + a} - \frac{c}{a + b}}$

  which shows that

  $\frac{a}{b + c} , \frac{b}{c + a}, \frac{c}{a + b}$ are in AP.

This completes the proof.